The horizontal and vertical displacements $x$ and $y$ of a projectile at a given time $t$ are given by $x=6 t$ metre and $y=8 t-5 t^2$ metre. The range of the projectile in metre is

Question

Aicharya · Accepted Answer

$ \begin{aligned} & x=(u \cos 	heta) t=6 t \ & \quad y=(u \sin 	heta) t-\frac{1}{2} g t^2=8 t-5 t^2 \end{aligned} $ Therefore, $u \sin 	heta=8$ $u \cos 	heta=6$ $\begin{aligned} Range\ R & =\frac{u^2 \sin 2 	heta}{g}=\frac{u^2 	imes 2 \sin 	heta \cos 	heta}{g} \ & =\frac{2(u \sin 	heta)(u \cos 	heta)}{g} \ & =\frac{2(8)(6)}{10}=9.6 \mathrm{~m}\end{aligned}$

The horizontal and vertical displacements $x$ and $y$ of a projectile at a given time $t$ are given by $x=6…

Solution